Past exam of the mathematics course of the year 2 of the natural sciences course of the University of Cambridge / 2016 / Paper 2 / 10

Since $a^2=I$, a representation satisfies $D(a{)}^2=I_2$. Orthogonality gives $D(a{)}^{-1}=D(a{)}^T$, while the involution relation gives $D(a{)}^{-1}=D(a)$. Hence $D(a)$ is a symmetric matrix, so $\gamma =\delta$.

Its entries obeyIf $\gamma \ne 0$, then $\beta =-\alpha$; putting $\alpha =\cos \theta$ and $\gamma =\sin \theta$ gives a reflection. If $\gamma =0$, the diagonal entries are independently $\pm 1$: the opposite-sign cases are already included among these reflections, while equal signs give $I_2$ or $-I_2$. Faithfulness excludes $I_2$, since $a\ne I$. Therefore the orthogonal involutions of the plane giveThe first is rotation by $\pi$. The second is reflection in the line making angle $\theta /2$ with the first axis: the unit vector along that line is fixed, and its perpendicular is negated. These are all the nonidentity orthogonal transformations of order two in the plane.

A cyclic group of order $p$ contains an element of order two if and only if $2$ divides $p$. Necessity follows from Lagrange's theorem; for even $p$, the unique such element is $c=g_1^{p/2}$. HenceBecause $g_1$ has order $p>2$ and the representation is faithful, $D(g_1)$ cannot be a reflection, which has order two. It is therefore a rotation through a primitive angle $\vartheta =2\pi q/p$, with $\gcd (q,p)=1$. The condition on the greatest common divisor is exactly what makes its order $p$ rather than a proper divisor. Since $p$ is even, $q$ is odd, so $D(c)$ is the half-turn:The character is the trace:Choosing the primitive generator with $q=1$ gives the customary rotation representation. The general coprime value is included because faithfulness alone does not select that generator. These are the two-dimensional cases in the irreducible real representations of a finite cyclic group.

Write $r=g_1$. The matrix $D(s)$ is a reflection by part (a). Reflection reverses a rotation angle, giving $D(s)D(r)D(s)=D(r{)}^{-1}$. Faithfulness lifts this matrix identity back to the group:Moving each $s$ past a rotation reduces every word to $r^i$ or $r^{i}s$, $0\le i<p$. These $2p$ elements are distinct: the rotations are distinct by the order of $r$, and reflections have determinant $-1$, so cannot equal a rotation. ThereforeHere the subscript denotes group order. This is the reflection case of finite subgroups of the plane orthogonal group.

Every $r^{i}s$ has square $r^{i}s{r}^{i}s=r^i{r}^{-i}=I$, so there are $p$ reflection subgroups of order two. Within the rotations, the even-$p$ assumption from part (b) supplies the additional half-turn $r^{p/2}$. Distinct involutions generate distinct two-element subgroups. ThusFor comparison, without the evenness requirement the corresponding odd-$p$ count would be $p$. The even case agrees with involutions in an even dihedral group.

For $p=4$, takeDirect multiplication gives $SRS=\left(\begin{array}{cc}0& 1\\ -1& 0\end{array}\right)=R^3$, so $srs=r^3=g_3$. The product rules arewith exponents modulo four. They yield the complete multiplication table, where the row element multiplies the column element on the left:This is the symmetry group of a square, the dihedral group of order eight, often written $D_4$ when its subscript counts vertices instead. Its five order-two subgroups are generated by $r^2,s,rs,r^{2}s,r^{3}s$.